2020廣東工業大學810自控原理

1、
胡壽松第一章課后習題第三題

2、
(1)
由系統的倍訓傳遞函式可知
ζ = 0.5 ω n = 2 \zeta=0.5 \quad \omega_n=2 ζ=0.5ωn?=2
零初始狀態下超調量
σ = e ? π ζ 1 ? ζ 2 × 100 % = 16.3 % \sigma=e^{\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}}\times100\%=16.3\% σ=e1?ζ2 ??πζ?×100%=16.3%
調節時間
t s = 4.4 ζ ω n = 4.4 s t_s=\frac{4.4}{\zeta\omega_n}=4.4s ts?=ζωn?4.4?=4.4s
(2)
考慮初始條件
Y ( s ) = 4 s 2 + 2 s + 4 R ( s ) + [ y ( 0 ) s + y ˙ ( 0 ) ] + 2 y ( 0 ) s 2 + 2 s + 4 = 4 s 2 + 2 s + 4 R ( s ) + 0.5 s + 1 s 2 + 2 s + 4 = 4 s ( s 2 + 2 s + 4 ) + 1 2 s + 1 ( s + 1 ) 2 + 3 + 1 6 3 ( s + 1 ) 2 + 3 \begin{aligned} Y(s)&=\frac{4}{s^2+2s+4}R(s)+\frac{[y(0)s+\dot{y}(0)]+2y(0)}{s^2+2s+4} \\ &=\frac{4}{s^2+2s+4}R(s)+\frac{0.5s+1}{s^2+2s+4} \\ &=\frac{4}{s(s^2+2s+4)}+\frac{1}{2}\frac{s+1}{(s+1)^2+3}+\frac{1}{6}\frac{3}{(s+1)^2+3} \end{aligned} Y(s)?=s2+2s+44?R(s)+s2+2s+4[y(0)s+y˙?(0)]+2y(0)?=s2+2s+44?R(s)+s2+2s+40.5s+1?=s(s2+2s+4)4?+21?(s+1)2+3s+1?+61?(s+1)2+33??
對上式取拉氏反變換
y ( t ) = 1 ? 2 3 3 e ? t sin ? ( 3 t + π 3 ) + 1 2 e ? t cos ? 3 t + 1 6 e ? t sin ? 3 t y(t)= 1-\frac{2\sqrt{3}}{3}e^{-t}\sin(\sqrt{3}t+\frac{\pi}{3})+\frac{1}{2}e^{-t}\cos\sqrt{3}t+\frac{1}{6}e^{-t}\sin\sqrt{3}t y(t)=1?323 ??e?tsin(3 ?t+3π?)+21?e?tcos3 ?t+61?e?tsin3 ?t
當 t = 1 t=1 t=1時
y ( 1 ) = 0.933 s y(1)=0.933s y(1)=0.933s

3、
(1)
系統的根軌跡方程
K s ( s + 4 ) ( s + 8 ) = ? 1 \frac{K}{s(s+4)(s+8)}=-1 s(s+4)(s+8)K?=?1
①根軌跡的起點
p 1 = 0 , p 2 = ? 4 , p 3 = ? 8 p_1=0,\quad p_2=-4,\quad p_3=-8 p1?=0,p2?=?4,p3?=?8
②根軌跡的終點為無窮遠處

③實軸上的根軌跡： ( ? ∞ , ? 8 ] , [ ? 4 , 0 ] (-\infin,-8],\quad [-4,0] (?∞,?8],[?4,0]

④根軌跡漸近線
0 ? 4 ? 8 3 = ? 4 , φ 1 = 60 ° , φ 2 = 300 ° \frac{0-4-8}{3}=-4, \quad \varphi_1=60°, \quad \varphi_2=300° 30?4?8?=?4,φ1?=60°,φ2?=300°
⑤根軌跡的分離點：
W ( s ) = s 3 + 12 s 2 + 32 s d W ( s ) d s = 3 s 2 + 24 s + 32 = 0 s 1 = ? 1.69 ( 分 離 點 ) , s 2 = ? 6.31 ( 舍 去 ) \begin{gathered} W(s)=s^3+12s^2+32s \\ \frac {dW(s)}{ds}=3s^2+24s+32=0 \\ s_1=-1.69(分離點),\quad s_2=-6.31(舍去) \end{gathered} W(s)=s3+12s2+32sdsdW(s)?=3s2+24s+32=0s1?=?1.69(分離點),s2?=?6.31(舍去)?
⑥根軌跡與虛軸的交點
系統的特征方程
D ( s ) = s 3 + 12 s 2 + 32 s + K = 0 D(s)=s^3+12s^2+32s+K=0 D(s)=s3+12s2+32s+K=0
令 s = j ω s=j\omega s=jω
D ( j ω ) = ? j ω 3 ? 12 ω 2 + 32 j ω + K = ( K ? 12 ω 2 ) + j ( 32 ω ? ω 3 ) = 0 \begin{aligned} D(j\omega)&=-j\omega ^3-12\omega ^2+32j\omega+K \\ &=(K-12\omega ^2)+j(32\omega-\omega ^3)=0 \end{aligned} D(jω)?=?jω3?12ω2+32jω+K=(K?12ω2)+j(32ω?ω3)=0?
令實部和虛部為零，可得
{ K = 384 ω = 5.66 \left\{\begin{array}{l} K = 384 \\ \omega = 5.66 \end{array}\right. {K=384ω=5.66?
圖略
(2)
由(1)可知，系統穩定時， 0 < K < 382 0<K<382 0<K<382
(3)
將分離點代入特征方程解得 K = 93.18 K=93.18 K=93.18，系統單位階躍回應無超調時， 0 < K < 93.18 0<K<93.18 0<K<93.18
(4)
設極點為 s 1 , 2 = ? a ± 3 a j s_{1,2}=-a\pm\sqrt{3}aj s1,2?=?a±3 ?aj，由根之和得 s 3 = 12 ? 2 a s_3=12-2a s3?=12?2a，則
D ( s ) = s 3 + 12 s 2 + 32 s + K = ( s + a + 3 a j ) ( s + a ? 3 a j ) ( s + 12 ? 2 a ) = s 3 + 12 s 2 + 2 a ( 12 ? 2 a ) s + 4 a 2 ( 12 ? 2 a ) \begin{aligned} D(s)&=s^3+12s^2+32s+K \\ &=(s+a+\sqrt{3}aj)(s+a-\sqrt{3}aj)(s+12-2a) \\ &=s^3+12s^2+2a(12-2a)s+4a^2(12-2a) \end{aligned} D(s)?=s3+12s2+32s+K=(s+a+3 ?aj)(s+a?3 ?aj)(s+12?2a)=s3+12s2+2a(12?2a)s+4a2(12?2a)?
比較系數解得
{ K = 83.52 a = 1.53 \left\{\begin{array}{l} K = 83.52 \\ a = 1.53 \end{array}\right. {K=83.52a=1.53?
極點分別為
s 1 , 2 = ? 1.53 ± 2.65 j , s 3 = ? 8.94 s_{1,2}=-1.53\pm2.65j,\quad s_3=-8.94 s1,2?=?1.53±2.65j,s3?=?8.94
第三個極點的模是共軛復數極點實部的模 5.8 5.8 5.8倍，這對共軛復數極點是倍訓系統的主導極點，降階后開環系統的傳遞函式為
G ( s ) = 9.57 s 2 + 3.06 s + 9.3694 G(s)=\frac{9.57}{s^2+3.06s+9.3694} G(s)=s2+3.06s+9.36949.57?
倍訓傳遞函式為
Φ ( s ) = 9.57 s 2 + 3.06 s + 18.9334 \varPhi(s)=\frac{9.57}{s^2+3.06s+18.9334} Φ(s)=s2+3.06s+18.93349.57?

4、
(1)
系統的頻率特性
G ( j ω ) = ? 0.5 ( ω 2 + 1 ) ( 0.25 ω 2 + 1 ) + j 0.5 ω 2 + 1 ω ( ω 2 + 1 ) ( 0.25 ω 2 + 1 ) G(j\omega)=-\frac{0.5}{(\omega^2+1)(0.25\omega^2+1)}+j\frac{0.5\omega^2+1}{\omega(\omega^2+1)(0.25\omega^2+1)} G(jω)=?(ω2+1)(0.25ω2+1)0.5?+jω(ω2+1)(0.25ω2+1)0.5ω2+1?
當 ω → 0 + \omega\rarr0^+ ω→0+時
R e [ G ( j ω ) ] = ? 0.5 , ∠ G ( j ω ) = ? 270 ° Re[G(j\omega)]=-0.5,\quad \angle G(j\omega)=-270° Re[G(jω)]=?0.5,∠G(jω)=?270°
奈奎斯曲線與負實軸無交點，圖略
(2)
由系統的開環傳遞函式可知， P = 1 P=1 P=1，正穿越 N ? = 0.5 N_-=0.5 N??=0.5，則
Z = P + 2 N ? = 2 Z=P+2N_-=2 Z=P+2N??=2
系統不穩定，倍訓系統在 s s s右半平面有2個極點
(3)
系統的特征方程
D ( s ) = s 3 + s 2 ? 2 s + 1 D(s)=s^3+s^2-2s+1 D(s)=s3+s2?2s+1
勞斯表
s 3 1 ? 2 s 2 1 1 s 1 ? 3 s 0 1 \begin{matrix} s^3& 1 & -2 & \\ s^2& 1 & 1 & \\ s^1& -3 & & \\ s^0& 1 \end{matrix} s3s2s1s0?11?31??21??
勞斯表第一列不均大于 0 0 0，且變號兩次，倍訓系統在 s s s右半平面有2個極點

5、
(1)
當系統的相角裕度為 γ = 30 ° \gamma=30° γ=30°
90 ° ? arctan ? ω c ? arctan ? 0.5 ω c = 30 ° 90°-\arctan\omega_c-\arctan0.5\omega_c=30° 90°?arctanωc??arctan0.5ωc?=30°
解得
ω c = 0.792 r a d / s \omega_c=0.792rad/s ωc?=0.792rad/s
代入系統的幅頻特性
10 K ω c ω c 2 + 1 0.25 ω c 2 + 1 = 1 \frac{10K}{\omega_c\sqrt{\omega_c^2+1}\sqrt{0.25\omega_c^2+1}}=1 ωc?ωc2?+1 ?0.25ωc2?+1 ?10K?=1
解得
K = 0.109 K=0.109 K=0.109
(2)
設滯后校正裝置的傳遞函式為
G c s = K c ( T s + 1 ) β T s + 1 G_c{s}=\frac{K_c(Ts+1)}{\beta Ts+1} Gc?s=βTs+1Kc?(Ts+1)?
輸入 r ( t ) = t r(t)=t r(t)=t時系統的穩態誤差為 e s s = 0.2 e_{ss}=0.2 ess?=0.2，則
K c = 0.5 K_c=0.5 Kc?=0.5
此時，系統的截止頻率為
5 ω c ω c 2 + 1 0.25 ω c 2 + 1 = 1 \frac{5}{\omega_c\sqrt{\omega_c^2+1}\sqrt{0.25\omega_c^2+1}}=1 ωc?ωc2?+1 ?0.25ωc2?+1 ?5?=1
解得
ω c = 1.802 r a d / s \omega_c=1.802rad/s ωc?=1.802rad/s
相角裕度為
γ = 180 ° ? 90 ° ? arctan ? ω c ? arctan ? 0.5 ω c = ? 12.991 ° \gamma=180°-90°-\arctan\omega_c-\arctan0.5\omega_c=-12.991° γ=180°?90°?arctanωc??arctan0.5ωc?=?12.991°
取相應的相角裕度
γ 1 = 46 ° \gamma_1=46° γ1?=46°
對應的截止頻率為
90 ° ? arctan ? ω c 1 ? arctan ? 0.5 ω c 1 = 46 ° 90°-\arctan\omega_{c1}-\arctan0.5\omega_{c1}=46° 90°?arctanωc1??arctan0.5ωc1?=46°
解得
ω c 1 = 0.547 r a d / s \omega_{c1}=0.547rad/s ωc1?=0.547rad/s
由 20 lg ? ∣ G 1 ( j ω c 1 ) ∣ = 20 lg ? β 20\lg|G_1(j\omega_{c1})|=20\lg\beta 20lg∣G1?(jωc1?)∣=20lgβ，解得
β = 7.735 \beta=7.735 β=7.735
選取 1 T = 0.2 ω c \frac{1}{T}=0.2\omega_c T1?=0.2ωc?， T = 9.14 T=9.14 T=9.14，則
G c s = 0.5 ( 9.14 s + 1 ) 70.6979 s + 1 G_c{s}=\frac{0.5(9.14s+1)}{70.6979s+1} Gc?s=70.6979s+10.5(9.14s+1)?

6、
(1)
系統的特征方程
D ( s ) = s 3 + 6 s 2 + ( 9 + K ) s + K a = 0 D(s)=s^3+6s^2+(9+K)s+Ka=0 D(s)=s3+6s2+(9+K)s+Ka=0
將極點 s 1 = ? 2 + 4 j s_1=-2+4j s1?=?2+4j代入上式，解得
{ K = 19 a = 40 19 \left\{\begin{array}{l} K = 19 \\ a = \frac{40}{19} \end{array}\right. {K=19a=1940??
由根之和解得
s 3 = ? 2 s_3=-2 s3?=?2
第三個極點與系統的倍訓零點組成偶極子，所以這對共軛復數極點是倍訓系統的主導極點，降階后系統的倍訓傳遞函式為
Φ ( s ) = 19 s 2 + 4 s + 20 \varPhi(s)=\frac{19}{s^2+4s+20} Φ(s)=s2+4s+2019?
由系統的倍訓傳遞函式可知
ζ = 5 5 ω n = 2 5 \zeta=\frac{\sqrt{5}}{5} \quad \omega_n=2\sqrt{5} ζ=55 ??ωn?=25 ?
零初始狀態下超調量
σ = e ? π ζ 1 ? ζ 2 × 100 % = 20.79 % \sigma=e^{\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}}\times100\%=20.79\% σ=e1?ζ2 ??πζ?×100%=20.79%
調節時間
t s = 4.4 ζ ω n = 2.2 s t_s=\frac{4.4}{\zeta\omega_n}=2.2s ts?=ζωn?4.4?=2.2s
(2)
當 G c = K p ′ G_c=K^{'}_p Gc?=Kp′?時，系統的特征方程為
D ( s ) = s 3 + 6 s 2 + 9 s + K = 0 D(s)=s^3+6s^2+9s+K=0 D(s)=s3+6s2+9s+K=0
令 s = j ω s=j\omega s=jω
D ( j ω ) = ? j ω 3 ? 6 ω 2 + 9 j ω + K = ( K ? 6 ω 2 ) + j ( 9 ω ? ω 3 ) = 0 \begin{aligned} D(j\omega)&=-j\omega ^3-6\omega ^2+9j\omega+K \\ &=(K-6\omega ^2)+j(9\omega-\omega ^3)=0 \end{aligned} D(jω)?=?jω3?6ω2+9jω+K=(K?6ω2)+j(9ω?ω3)=0?
令實部和虛部為零，可得
{ K = 54 ω = 3 r a d / s \left\{\begin{array}{l} K = 54 \\ \omega = 3 rad/s \end{array}\right. {K=54ω=3rad/s?
等幅振蕩周期為
P = 2 π ω = 2.09 P=\frac{2\pi}{\omega}=2.09 P=ω2π?=2.09
根據表可得
K p = 32.4. T i = 1.045 , T d = 0.26125 K_p=32.4.\quad T_i=1.045,\quad T_d=0.26125 Kp?=32.4.Ti?=1.045,Td?=0.26125
引入比例控制，根據表可得系統的開環傳遞函式為
G 1 ( s ) = 27 s ( s + 3 ) 2 G_1(s)=\frac{27}{s(s+3)^2} G1?(s)=s(s+3)227?
比例控制可以提高系統的開環增益，使穩態誤差減少，但是 K p K_p Kp?過大，會影響系統的穩定性，導致系統不穩定，
再引入積分控制，根據表可得系統的開環傳遞函式為
G 2 ( s ) = 24.3 1.74097 1.74097 s + 1 s 2 ( s + 3 ) 2 G_2(s)=\frac{24.3}{1.74097}\frac{1.74097s+1}{s^2(s+3)^2} G2?(s)=1.7409724.3?s2(s+3)21.74097s+1?
積分控制可以提高系統的型別，消除或減少穩態誤差，使系統的穩態性能得到改善，但穩定裕度減少，同時還引進一個開環零點，改善系統的動態性能，
最后引入微分控制，根據表可得系統的開環傳遞函式為
G 3 ( s ) = 32.4 1.045 0.27300625 s 2 + 1.045 s + 1 s 2 ( s + 3 ) 2 G_3(s)=\frac{32.4}{1.045}\frac{0.27300625s^2+1.045s+1}{s^2(s+3)^2} G3?(s)=1.04532.4?s2(s+3)20.27300625s2+1.045s+1?
微分控制給系統增加一個零點，提高系統的回應速度，改善系統的動態性能，

7、
由于 M > h M>h M>h，兩個非線性特性串聯得到一個理想繼電特性，其中 M = 1 M=1 M=1，圖略，系統線性部分的幅值為
ω x = 1 T 1 T 2 = 3 3 , G ( j ω x ) = ? K T 1 T 2 T 1 + T 2 = ? 1 \omega_x=\frac{1}{\sqrt{T_1T_2}}=\frac{\sqrt{3}}{3},\quad G(j\omega_x)=\frac{-KT_1T_2}{T_1+T_2}=-1 ωx?=T1?T2? ?1?=33 ??,G(jωx?)=T1?+T2??KT1?T2??=?1
非線性部分
? 1 N ( A ) = ? π A 4 , ? 1 N ( 0 ) = 0 , ? 1 N ( ∞ ) = ? ∞ -\frac{1}{N(A)}=-\frac{\pi A}{4},\quad -\frac{1}{N(0)}=0,\quad -\frac{1}{N(\infin)}=-\infin ?N(A)1?=?4πA?,?N(0)1?=0,?N(∞)1?=?∞
圖略， Γ G \Gamma_G ΓG?與 ? 1 / N ( A ) -1/N(A) ?1/N(A)的交點為 ( ? 1 , 0 ) (-1,0) (?1,0)， ? 1 / N ( A ) -1/N(A) ?1/N(A)曲線沿著 A A A增大的方向由不穩定區域進入穩定區域，該點的周期運動是穩定的，對應的自振頻率為
ω = 3 3 \omega=\frac{\sqrt{3}}{3} ω=33 ??
由 R e [ G ( j ω ) N ( A ) ] = ? 1 Re[G(j\omega)N(A)]=-1 Re[G(jω)N(A)]=?1可解得自振振幅為
A = 4 π A=\frac{4}{\pi} A=π4?

8、
(1)
系統的脈沖傳遞函式
G ( z ) = ( 1 ? z ? 1 ) Z [ K s ( s + 1 ) ] = 0.632 K z ? 0.368 G(z)=(1-z^{-1})Z[\frac{K}{s(s+1)}]=\frac{0.632K}{z-0.368} G(z)=(1?z?1)Z[s(s+1)K?]=z?0.3680.632K?
系統的特征方程
D ( z ) = ( 0.632 K + 0.632 ) z + ( 1.368 ? 0.632 K ) D(z)=(0.632K+0.632)z+(1.368-0.632K) D(z)=(0.632K+0.632)z+(1.368?0.632K)
對上式進行雙線變換
D ( w ) = ? K w + 3 K + 2 D(w)=-Kw+3K+2 D(w)=?Kw+3K+2
系數大于零，系統穩定，則 ? 2 3 < K < 2.164 -\frac{2}{3}<K<2.164 ?32?<K<2.164
(2)
當 K = 1 , r ( t ) = 1 ( t ) K=1,\quad r(t)=1(t) K=1,r(t)=1(t)時，系統輸出為
Y ( z ) = 0.632 z ( z ? 1 ) ( z + 0.264 ) Y(z)=\frac{0.632z}{(z-1)(z+0.264)} Y(z)=(z?1)(z+0.264)0.632z?
長除法可得
Y ( z ) = 0.632 z ? 1 + 0.465122 z ? 2 + 0.34232832 z ? 3 + ? Y(z)=0.632z^{-1}+0.465122z^{-2}+0.34232832z^{-3}+\cdots Y(z)=0.632z?1+0.465122z?2+0.34232832z?3+?
對上式進行變換
y ( k ) = 0.632 y ( k ? 1 ) + 0.465122 y ( k ? 2 ) + 0.34232832 y ( k ? 3 ) + ? y(k)=0.632y(k-1)+0.465122y(k-2)+0.34232832y(k-3)+\cdots y(k)=0.632y(k?1)+0.465122y(k?2)+0.34232832y(k?3)+?
或者由系統的倍訓傳遞函式可得
( z + 0.264 ) Y ( z ) = 0.632 R ( z ) (z+0.264)Y(z)=0.632R(z) (z+0.264)Y(z)=0.632R(z)
對上式進行變換
0.264 y ( k ) = ? y ( k ? 1 ) + 0.632 ? 1 ( t ) 0.264y(k)=-y(k-1)+0.632\cdot 1(t) 0.264y(k)=?y(k?1)+0.632?1(t)
(3)
系統的誤差傳遞函式為
Φ e ( z ) = z ? 0.368 z + 0.632 K ? 0.368 \varPhi_e(z)=\frac{z-0.368}{z+0.632K-0.368} Φe?(z)=z+0.632K?0.368z?0.368?
由終值定理可得系統的穩態誤差為
e s s ( ∞ ) = lim ? z → 1 = ( z ? 1 ) R ( z ) Φ e ( z ) = z ( z ? 0.368 ) z + 0.632 K ? 0.368 = 0.632 + 0.632 K ? 0.368 e_{ss}(\infin)=\lim\limits_{z\rightarrow1}=(z-1)R(z)\varPhi_e(z)=\frac{z(z-0.368)}{z+0.632K-0.368}=\frac{0.632}{+0.632K-0.368} ess?(∞)=z→1lim?=(z?1)R(z)Φe?(z)=z+0.632K?0.368z(z?0.368)?=+0.632K?0.3680.632?